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1 year ago

15

Which reactant is unlikely to produce the indicated product upon strong heating?

1 answer:

Ilya [14]1 year ago

8 0

2-Methyl-4-oxo-pentanoic acid is unlikely to produce 2-Methyl-3-butanone upon strong heating.

Upon heating, the β ketoacid becomes unstable and decarboxylates, leading to the formation of the methyl ketone.

A carboxylic acid is an organic acid that contains a carboxyl group (C(=O)OH) attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO2H, with R referring to the alkyl, alkenyl, aryl, or other group.

Carboxylic acids occur widely. Important examples include the amino acids and fatty acids. Deprotonation of a carboxylic acid gives a carboxylate anion.

Full question :

Q. Which reactant is unlikely to produce the indicated product upon strong heating?

A) 2,2-Dimethylpropanedioic acid 2-methylpropanoic acid
B) 2-Ethylpropanedioic acid Butanoic acid
C) 2-Methyl-3-oxo-pentanoic acid 3-Pentanone
D) 2-Methyl-4-oxo-pentanoic acid 2-Methyl-3-butanone
E) 4-Methyl-3-oxo-heptanoic acid 3-Methyl-2-hexanone

Hence, option (D) is correct.

Learn more about carboxylic acid here : brainly.com/question/26855500

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Co(g) effuses at a rate that is ______ times that of cl2(g) under the same conditions.

il63 [147K]

Answer is: the ratio of the effusion rate is 1.59 : 1.

1) rate of effusion of carbon monoxide gas = 1/√M(CO).

rate of effusion of carbon monoxide gas = 1/√28.

rate of effusion of carbon monoxide gas = 0.189.

2) rate of effusion of chlorine = 1/√M(Cl₂).

rate of effusion of chlorine = 1/√70.9.

rate of effusion of chlorine = 0.119.

rate of effusion of carbon monoxide : rate of effusion of chlorine =

= 0.189 : 0.119 / ÷0.119.

rate of effusion of carbon monoxide : rate of effusion of chlorine = 1.59 : 1.

4 0

3 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

3 0

3 years ago

Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?

dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

C₄H₁₀: 2 moles
O₂: 13 moles
CO₂: 8 moles
H₂O: 10 moles

Being:

C: 12 g/mole
H: 1 g/mole
O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
O₂: 2*16 g/mole= 32 g/mole
CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

C₄H₁₀: 2 moles* 58 g/mole= 116 g
O₂: 13 moles* 32 g/mole= 416 g
CO₂: 8 moles* 44 g/mole= 352 g
H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

$mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10} }{116 grams of C_{4}H_{10}}$

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

3 0

2 years ago

The breaking and forming of bonds between atoms in substances results in changes in the _______ contained in the substances.

photoshop1234 [79]

Answer:

Electrons

Explanation:

The breaking and forming of bonds between atoms in substances results in changes in the number of electrons in the substance.

Atoms combined in order to share, gain or lose electrons for it to be stable.
The noble gases have a set up configuration which makes them stable.
All atom tend to mimic the noble gases.

6 0

2 years ago

A chemist prepares a solution of silver nitrate by measuring out of silver nitrate into a volumetric flask and filling the flask

Snezhnost [94]

Amount of silver nitrate taken = 269.μmol $AgNO_{3}$

Volume of the solution = 300. mL

Concentration of a solution is generally expressed in terms of molarity. Molarity is defined as the moles of a substance present per liter of the solution.

$Molarity = \frac{Moles of solute}{Volume of solution(L)}$

We want the concentration in millimoles/L.

Converting μmol to millimol solute:

$269.$ μ $mol * \frac{1 millimol}{1000 micromol}$ = 0.269 millimol

Volume from mL to L: $300. mL * \frac{1 L}{1000 mL} = 0.300 L$

Therefore concentration of the chemist's solution = $\frac{0.269 millimol}{0.300 L} = 0.897 \frac{millimol}{L}$

8 0

3 years ago